<PRE><B>Question 276055</B>
A farmer spends $4,000 to obtain 100 head of livestock.Prices are: calves – $120 each; lambs – $50 each; piglets – $25 each.
If he purchased at least one animal of each type and fewer than 10 calves,how many lambs did he buy?
;
Total animal equation:
C + L + P = 100
:
Total cost equation
120C + 50L + 25P = 4000
Simplify, divide by 5
24C + 10L + 5P = 800
:
Multiply the 1st equation by 5 subtract from the above, get rid of the pigs!
24C + 10L + 5P = 800
5C  + 5L  + 5P = 500
-----------------------Subtraction, eliminates p
19C + 5L = 300
5L = 300 - 19C
L = {{{300/5}}} - {{{19/5}}}C
L = 60 - {{{19/5}}}C
:
We know L has to be an integer, therefore C has to be a multiple of 5
they tell his their are fewer than 10 Calves so obviously C = 5
L = 60 - {{{19/5}}}5
L = 60 - 19
L = 41 lambs
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Find the pigs
100 - 5 - 41 = 54 pigs
:
:
Check solution by finding the total cost
120(5) + 41(50) + 25(54) = 
600 + 2050 + 1350 = 4000
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